STRUCTURE OF F19, F18 AND O18
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks ,discovered by Gell-Mann and Zweig, I published my paper "Nuclear structure is governed by the fundamental laws of electromagnetism " (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure. Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Nuclear structure of the stable F-19 with S = +1/2 In the following diagram of the structure of F19 and F18 one sees that the total spins are the fundamental characteristics which lead to the nuclear structure of F19 and F18. In Fig. 8d of my published paper you see that the structure of F-19 is based on the very stable structure of O16 with S = 0 which plays the role of a core with the fundamental four horizontal squares. The nucleons of the second and third square form a parallelepiped of a great stability because it is characterized by points with four pn bonds per nucleon. This is the central parallelepiped of the structure of O-16 which is responsible for the stability of O-16 with B(O16) = -127.58 MeV. In the following diagram of F19 you see that the extra p9 of the first square and n9 of the second square respectively form a stable rectangle of S = 0 which makes a blank position at n10 able to receive the extra n10 with S = +1/2 of the third square giving the total spin S = +1/2 of the F19. Such an extra neutron forms the two stable bonds n10p9 and n10p6 responsible for the stability of F19 with B(F-19) = -143.16 MeV Thus B(F-19 ) – B(O-16) = - 15.58 MeV This means that the three extra radial and the two extra axial bonds of the extra rectangle overcome the extra repulsions of the pp and nn systems. Stable F19 with S = +1/2 Unstable F18 with S = +1 p8(-1/2).n8(-1/2) p8(-1/2).n8(-1/2) n7(-1/2).p7(-1/2) n7(-1/2).p7(-1/2) n6(+1/2).p6(+1/2)….n10(+1/2) p9(+1/2)…. n6(+1/2).p6(+1/2)....n9(+1/2) p5(+1/2).n5(+1/2) p5(+1/2).n5(+1/2) p4(-1/2).n4(-1/2)….p9(-1/2) p4(-1/2).n4(-1/2) n3 (-1/2).p3 (- 1/2). n3(-1/2).p3(-1/2) n2(+1/2).p2(+1/2)….n9(+1/2) n2(+1/2).p2(+1/2) p1(+1/2). n1(+1/2) p1(+1/2).n1(+1/2) Nuclear structure of the unstable F18 with S = +1 Here both p9 and n9 form the single bonds of p9n6 and n9p6 of the third horizontal square respectively. Surprisingly the single n9p6 is stable because at point p6 there exist five pn bonds per nucleon which contribute to the increase of the binding energy of the extra n9p6 bond. However the p9n6 single bond is unstable because all protons of the core (Oxygen) exert repulsive forces of long range on p9 with a great repulsive energy Q. Therefore p9 turns into n10 as p9 + Q = n10 + positron + neutrino Such a decay leads to the structure of the stable O-18. ' ' Nuclear structure of the stable O18 with S = 0 After the transformation of the p9 into the neutron n10, one concludes that at the same time the neutron n10 with a spin -1/2 goes to the second horizontal square and forms the single bond n10p4. Thus the spin of the O18 is S = 0, because the two extra neutrons of n9 and n10 have opposite spin. It is of interest to note that both single bonds of n9p6 and n10p4 having the structure of deuterons are stable, because at points p6 and p4 respectively there exist five pn bonds per nucleon which contribute to the increase of the binding energies of the extra single bonds. Note that when the binding energy of a single np bond is stronger than the value of 1.29 MeV it cannot lead to the decay because the difference in energies between the down quark (3.69 MeV) and the up quark (2.4 MeV) is 1.29 MeV. Category:Fundamental physics concepts